Sortir Daftar Dengan Swap dan Pop

21

Pertimbangkan daftar bilangan bulat acak 1 hingga N. Anda ingin mengurutkannya hanya menggunakan tindakan berikut:

  1. Tukar elemen daftar pertama dan terakhir. (S)
  2. Pop off elemen pertama dan menambahkan ke akhir daftar. (P)

Ini selalu dimungkinkan karena daftar apa pun dapat diurutkan dengan pertukaran elemen-elemen yang berdekatan. Menggunakan S dan P Anda dapat menukar elemen yang berdekatan dengan memanggil P hingga dua elemen yang dimaksud adalah item pertama dan terakhir dalam daftar, lalu memanggil S untuk menukar mereka, lalu memanggil P lagi sampai mereka berada dalam indeks asli (sekarang ditukar ).

Namun, dalam hal jumlah operasi S dan P, metode ini tidak mungkin optimal untuk sebagian besar daftar.

Tantangan

Tulis program atau fungsi yang memperhitungkan permutasi angka dari 1 hingga N (N> 1). Itu dapat diberikan sebagai daftar atau string atau apa pun yang nyaman. Seharusnya output urutan S dan P yang akan mengurutkan permutasi ketika diterapkan dari kiri ke kanan. Urutan ini tidak harus pendek secara optimal, tetapi lebih pendek itu lebih baik (lihat Penilaian).

Contoh

Jika input itu [2, 1, 3]output mungkin SPkarena

  • S diterapkan untuk [2, 1, 3]membuatnya [3, 1, 2],
  • dan P diterapkan untuk [3, 1, 2]membuatnya [1, 2, 3], yang diurutkan.

Penguji

Cuplikan tumpukan ini dapat digunakan untuk memverifikasi bahwa urutan benar-benar mengurutkan daftar. Daftar harus memiliki tanda kurung siku dan dipisahkan koma. Urutan SP hanyalah string S's dan P' s.

<style>*{font-family:monospace}</style><script>function swap(list) {var tmp = list[0];list[0] = list[list.length - 1];list[list.length - 1] = tmp;}function pop(list) {list.push(list.shift())}function check() {var result = 'Sorted sucessfully.';var details = '';try {var list = eval(document.getElementById('list').value);var seq = document.getElementById('seq').value;details += 'Sequence: ' + seq + '<br>Steps:<br>' + list + ' <-- original';for(var i = 0; i < seq.length; i++) {if (seq.charAt(i) == 'S')swap(list);else if (seq.charAt(i) == 'P')pop(list);details += ' (' + i + ')<br>' + list;}details += ' <-- final (' + i + ')';for(var i = 1; i < list.length; i++) {if (list[i-1] > list[i]) {result = 'Sorting failed!';break;}}} catch(e) {result = 'Error: ' + e;}document.getElementById('result').innerHTML = result;document.getElementById('details').innerHTML = details;}</script><p>List<br><input id='list'type='text'size='60'value='[2, 1, 3]'></p><p>SP Sequence<br><textarea id='seq'rows='1'cols='60'>SP</textarea><br></p><button type='button'onclick='check()'>Check</button><p id='result'></p><p id='details'></p>

Mencetak gol

Algoritme Anda harus menghasilkan urutan SP yang benar, tetapi mungkin kurang optimal, untuk N = 2 hingga 256. Untuk permutasi apa pun dalam rentang ini, ia harus berjalan dalam waktu kurang dari 5 menit pada komputer modern yang layak .

Skor Anda adalah jumlah total operasi S dan P algoritma Anda perlu mengurutkan semua 30 daftar yang disediakan di bawah ini. Skor terendah menang.

Anda tidak boleh membuat hardcode program Anda agar sesuai dengan data ini. Jika tampaknya perlu, saya dapat menambahkan lebih banyak data uji untuk menentukan pemenang.

1. Length 3
[2, 1, 3]
2. Length 7
[2, 7, 5, 3, 4, 6, 1]
3. Length 41
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4. Length 52
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5. Length 65
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6. Length 69
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7. Length 75
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8. Length 80
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9. Length 103
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10. Length 108
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11. Length 109
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12. Length 151
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13. Length 173
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14. Length 177
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15. Length 192
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16. Length 201
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17. Length 201
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18. Length 205
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19. Length 211
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20. Length 226
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21. Length 227
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22. Length 230
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23. Length 238
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24. Length 241
[2, 33, 49, 83, 207, 204, 13, 57, 115, 86, 102, 219, 232, 44, 177, 197, 171, 227, 191, 10, 25, 162, 62, 11, 76, 214, 163, 201, 130, 91, 233, 194, 112, 179, 66, 139, 183, 116, 196, 193, 150, 211, 30, 144, 209, 97, 174, 3, 68, 38, 120, 165, 56, 64, 87, 15, 79, 131, 206, 96, 188, 7, 99, 195, 129, 8, 186, 78, 212, 125, 161, 230, 225, 239, 47, 107, 53, 218, 164, 106, 198, 215, 181, 226, 6, 175, 167, 236, 67, 80, 210, 128, 155, 40, 63, 74, 113, 89, 18, 190, 124, 221, 59, 149, 103, 42, 156, 157, 200, 168, 34, 77, 65, 146, 5, 187, 222, 231, 140, 141, 172, 234, 100, 94, 132, 237, 24, 216, 152, 22, 51, 69, 35, 43, 105, 23, 61, 1, 72, 135, 104, 9, 12, 21, 46, 192, 159, 205, 158, 109, 28, 98, 50, 122, 111, 71, 166, 229, 37, 114, 173, 134, 136, 81, 121, 185, 118, 223, 20, 14, 108, 82, 178, 54, 26, 153, 36, 39, 117, 147, 95, 90, 16, 17, 170, 119, 199, 19, 84, 213, 88, 93, 151, 4, 101, 142, 110, 184, 55, 138, 75, 133, 148, 145, 45, 70, 217, 143, 224, 241, 31, 240, 182, 52, 238, 126, 85, 154, 137, 160, 27, 180, 60, 29, 32, 169, 235, 123, 176, 48, 220, 203, 228, 127, 41, 58, 73, 92, 189, 202, 208]
25. Length 241
[105, 160, 132, 32, 10, 117, 76, 2, 190, 108, 178, 51, 71, 237, 232, 47, 14, 124, 100, 31, 169, 196, 8, 184, 21, 151, 223, 86, 42, 127, 55, 58, 229, 4, 219, 46, 238, 179, 24, 194, 203, 122, 66, 15, 81, 146, 172, 106, 129, 135, 216, 120, 92, 231, 144, 195, 181, 162, 69, 45, 137, 136, 9, 30, 5, 188, 91, 49, 147, 233, 198, 17, 241, 163, 36, 18, 183, 59, 16, 29, 116, 182, 41, 48, 23, 39, 154, 210, 68, 167, 95, 213, 79, 225, 37, 157, 109, 143, 78, 142, 173, 155, 200, 110, 20, 73, 141, 168, 156, 126, 150, 201, 114, 1, 230, 211, 217, 131, 140, 204, 209, 149, 103, 199, 165, 175, 35, 107, 74, 63, 193, 239, 218, 234, 197, 224, 174, 121, 60, 88, 22, 171, 133, 207, 152, 34, 43, 228, 125, 115, 101, 7, 12, 220, 82, 153, 134, 52, 130, 70, 72, 44, 177, 89, 65, 98, 94, 53, 208, 227, 161, 3, 123, 236, 221, 87, 102, 50, 90, 180, 185, 186, 67, 77, 26, 212, 27, 222, 6, 145, 11, 13, 84, 25, 164, 64, 85, 139, 214, 189, 61, 96, 191, 128, 93, 138, 148, 19, 119, 202, 75, 176, 159, 56, 104, 206, 40, 187, 215, 62, 166, 240, 112, 226, 170, 83, 97, 57, 99, 38, 28, 113, 205, 158, 235, 111, 54, 192, 118, 80, 33]
26. Length 244
[89, 13, 154, 161, 235, 225, 92, 188, 215, 194, 54, 58, 128, 21, 165, 85, 144, 205, 142, 77, 109, 24, 83, 69, 72, 7, 224, 240, 191, 204, 183, 203, 68, 70, 63, 95, 206, 170, 153, 180, 45, 178, 35, 27, 190, 132, 222, 41, 40, 156, 97, 20, 217, 177, 167, 65, 23, 136, 216, 234, 38, 201, 236, 164, 82, 241, 10, 141, 148, 229, 5, 125, 113, 159, 193, 187, 130, 179, 52, 108, 86, 196, 174, 123, 56, 116, 227, 30, 239, 98, 186, 67, 135, 118, 163, 43, 32, 50, 231, 226, 232, 172, 200, 99, 6, 143, 39, 93, 107, 34, 129, 157, 100, 127, 15, 84, 81, 73, 121, 220, 44, 8, 57, 105, 91, 171, 162, 211, 244, 104, 112, 238, 212, 133, 71, 192, 145, 160, 87, 181, 60, 184, 119, 4, 55, 96, 53, 12, 213, 124, 94, 22, 42, 3, 48, 131, 117, 140, 138, 228, 219, 155, 59, 29, 202, 169, 114, 101, 47, 233, 176, 139, 207, 33, 79, 16, 51, 66, 75, 90, 198, 168, 166, 31, 151, 49, 208, 150, 111, 14, 25, 197, 17, 76, 80, 78, 9, 173, 26, 137, 19, 120, 199, 106, 152, 88, 147, 36, 158, 28, 243, 221, 110, 74, 175, 237, 61, 2, 62, 214, 189, 134, 195, 218, 18, 102, 46, 103, 11, 122, 146, 185, 182, 209, 1, 149, 115, 64, 37, 126, 230, 223, 242, 210]
27. Length 246
[28, 186, 214, 18, 220, 73, 20, 88, 234, 187, 27, 102, 22, 129, 10, 228, 196, 56, 47, 2, 16, 67, 124, 137, 177, 179, 223, 147, 188, 23, 103, 109, 149, 60, 229, 99, 222, 90, 49, 80, 158, 93, 171, 71, 175, 143, 19, 212, 40, 226, 219, 120, 146, 66, 167, 232, 94, 174, 237, 9, 173, 70, 122, 241, 58, 82, 191, 211, 180, 104, 53, 36, 83, 37, 131, 25, 162, 32, 210, 144, 145, 69, 135, 63, 154, 165, 46, 57, 50, 74, 128, 140, 112, 118, 125, 231, 221, 233, 95, 200, 153, 6, 166, 98, 208, 91, 89, 141, 4, 26, 134, 86, 8, 30, 157, 156, 224, 201, 243, 7, 161, 1, 84, 115, 44, 230, 78, 79, 5, 17, 194, 148, 152, 121, 193, 107, 240, 181, 29, 43, 65, 164, 45, 110, 64, 195, 216, 127, 59, 197, 178, 151, 139, 85, 75, 11, 204, 184, 77, 54, 51, 205, 108, 142, 130, 138, 238, 87, 38, 101, 96, 31, 215, 244, 242, 172, 213, 136, 97, 35, 39, 76, 42, 245, 123, 227, 24, 168, 33, 159, 217, 239, 55, 176, 68, 207, 106, 132, 15, 116, 61, 198, 62, 182, 225, 202, 105, 150, 183, 81, 170, 13, 41, 199, 3, 185, 235, 14, 155, 21, 126, 119, 169, 72, 12, 236, 48, 209, 190, 113, 218, 100, 52, 111, 189, 92, 192, 114, 117, 160, 133, 203, 163, 34, 206, 246]
28. Length 250
[119, 57, 59, 181, 212, 120, 236, 121, 99, 247, 138, 187, 175, 108, 107, 197, 123, 101, 141, 77, 201, 3, 52, 60, 56, 240, 157, 39, 42, 199, 23, 18, 136, 74, 137, 143, 229, 170, 20, 160, 206, 219, 191, 185, 46, 223, 150, 190, 116, 96, 198, 221, 220, 159, 238, 48, 176, 113, 168, 33, 44, 142, 67, 244, 13, 218, 122, 246, 214, 234, 237, 27, 165, 24, 153, 90, 84, 154, 235, 196, 80, 111, 102, 231, 228, 135, 16, 148, 26, 45, 10, 71, 156, 224, 183, 232, 72, 94, 132, 54, 58, 248, 144, 213, 139, 129, 245, 115, 25, 164, 87, 19, 193, 81, 32, 78, 91, 167, 171, 40, 92, 226, 109, 69, 86, 38, 61, 5, 14, 118, 145, 103, 53, 93, 172, 178, 225, 68, 163, 210, 179, 192, 208, 169, 17, 12, 50, 233, 6, 55, 243, 158, 88, 188, 242, 36, 173, 126, 155, 184, 216, 149, 47, 76, 200, 1, 112, 28, 161, 174, 41, 73, 222, 66, 37, 63, 64, 124, 89, 205, 9, 186, 202, 70, 203, 127, 105, 250, 182, 79, 43, 133, 8, 241, 114, 128, 51, 83, 98, 49, 209, 7, 95, 151, 162, 189, 180, 75, 195, 62, 207, 21, 104, 30, 117, 110, 140, 29, 227, 249, 82, 152, 11, 34, 85, 106, 217, 211, 2, 35, 215, 4, 230, 134, 31, 194, 97, 22, 125, 130, 100, 239, 131, 166, 65, 15, 146, 177, 204, 147]
29. Length 253
[46, 245, 174, 180, 85, 29, 141, 70, 252, 119, 214, 225, 86, 91, 41, 67, 219, 118, 127, 243, 2, 71, 157, 114, 55, 92, 5, 200, 199, 139, 191, 235, 153, 102, 206, 171, 117, 58, 223, 249, 11, 211, 202, 175, 156, 133, 57, 163, 47, 65, 213, 247, 189, 111, 177, 49, 124, 154, 164, 145, 80, 15, 232, 142, 39, 69, 201, 185, 229, 215, 170, 42, 3, 248, 169, 136, 149, 218, 237, 90, 135, 7, 242, 246, 23, 186, 31, 4, 167, 207, 173, 132, 195, 196, 78, 212, 22, 35, 194, 54, 184, 183, 222, 230, 88, 43, 144, 151, 34, 97, 6, 109, 37, 147, 72, 158, 134, 33, 51, 238, 165, 162, 9, 63, 166, 113, 137, 198, 50, 121, 106, 224, 19, 104, 95, 129, 193, 79, 192, 122, 30, 12, 234, 168, 76, 103, 73, 66, 188, 178, 172, 176, 250, 182, 110, 231, 155, 26, 197, 10, 152, 98, 131, 25, 36, 81, 138, 150, 227, 24, 112, 120, 204, 75, 8, 179, 94, 17, 140, 32, 77, 61, 233, 38, 205, 93, 99, 27, 208, 56, 216, 48, 241, 20, 130, 240, 115, 83, 60, 108, 28, 13, 89, 128, 160, 82, 40, 126, 16, 253, 21, 251, 228, 59, 159, 64, 203, 236, 52, 18, 181, 105, 107, 239, 220, 44, 74, 125, 209, 84, 226, 217, 210, 190, 101, 100, 68, 116, 161, 148, 244, 221, 96, 45, 123, 187, 62, 87, 53, 1, 146, 143, 14]
30. Length 256
[159, 248, 75, 43, 111, 38, 4, 17, 155, 87, 81, 208, 53, 230, 65, 11, 108, 228, 146, 212, 137, 225, 144, 189, 86, 105, 84, 128, 97, 50, 223, 15, 83, 169, 217, 47, 88, 236, 114, 181, 115, 177, 102, 250, 246, 104, 80, 45, 240, 14, 196, 52, 247, 41, 198, 32, 182, 206, 226, 101, 70, 94, 113, 49, 254, 59, 42, 154, 77, 253, 112, 215, 99, 25, 134, 92, 95, 150, 64, 178, 118, 79, 130, 63, 129, 131, 57, 218, 85, 204, 3, 163, 158, 186, 10, 199, 24, 125, 251, 51, 74, 160, 207, 120, 233, 30, 19, 9, 56, 167, 58, 117, 164, 54, 220, 229, 234, 203, 211, 103, 205, 69, 39, 5, 31, 191, 106, 180, 116, 245, 21, 222, 91, 235, 8, 2, 214, 29, 238, 176, 135, 61, 227, 202, 122, 252, 231, 89, 187, 37, 149, 96, 190, 172, 73, 18, 71, 1, 179, 46, 156, 201, 165, 256, 151, 27, 242, 188, 126, 124, 244, 100, 107, 143, 170, 72, 255, 40, 67, 192, 185, 219, 20, 157, 98, 237, 136, 168, 141, 224, 232, 44, 139, 132, 22, 60, 109, 90, 175, 171, 62, 23, 161, 12, 76, 210, 68, 184, 93, 243, 48, 209, 174, 200, 121, 123, 36, 173, 140, 133, 145, 6, 110, 152, 142, 241, 55, 138, 147, 193, 213, 127, 7, 34, 66, 153, 26, 13, 162, 183, 239, 78, 249, 221, 28, 194, 16, 35, 33, 82, 195, 148, 216, 119, 166, 197]
Hobi Calvin
sumber
Ini tantangan yang menarik. Tetapi pertanyaannya adalah: apakah kita harus menghitung skor secara manual, atau akankah Anda melakukannya di papan peringkat?
Pengoptimal
1
@Optimizer Saya lebih suka Anda menghitung skor Anda, jadi saya tidak perlu repot untuk setiap jawaban atau edit baru. Saya akan menjalankan program untuk mengonfirmasi skor yang lebih baik sendiri ketika memilih pemenang.
Calvin Hobi
Berikut adalah beberapa kasus uji ekstra: [2,1], [2,1,4,3], [1,2,6,4,5,3]. Kalau-kalau ada orang yang mencoba hal-hal rumit :)
Sp3000

Jawaban:

9

Python 3 (dengan PyPy) - skor: 607771

Sunting: Saya pikir bug sudah diperbaiki, tapi saya masih tidak yakin ini berfungsi sampai saya membuktikannya

def d(a, b, N):
    a, b = a%N, b%N

    if a >= b:
        return a-b
    else:
        return a+(N-b)

def gt(a, b, N):
    if a == N and b == 1:
        return False
    if a == 1 and b == N:
        return True
    return a > b

def solve(numbers):
    N = len(numbers)
    best_move = None
    target = sorted(numbers)

    if N == 2:
      return "" if numbers == [1,2] else "P"

    for b in range(N):
        nums = numbers[:]
        moves = []
        head_ptr = 0 # Which element should be at the start of the list if we actually moved stuff
        reference = target[b:]+target[:b]
        ref_d = {t:i for i,t in enumerate(reference)}

        while True:
            for pops in range(N):
                ptr = (head_ptr + pops) % N

                # Magic
                d1 = d(ref_d[nums[ptr-1]],ptr-1,N)+d(ptr,ref_d[nums[ptr]],N)
                d2 = d(ref_d[nums[ptr]],ptr,N)+d(ptr-1,ref_d[nums[ptr-1]],N)

                if d1 > d2 or (d1 == d2 and gt(nums[ptr-1], nums[ptr], N)):
                    moves.append("P"*pops + "S")
                    head_ptr = (head_ptr + pops) % N
                    nums[ptr-1],nums[ptr] = nums[ptr],nums[ptr-1]
                    break

            else:
                if nums[nums.index(1):]+nums[:nums.index(1)] != target:
                    print("This algorithm is seriously broken.\n")
                moves.append("P"*((N+nums.index(1)-head_ptr)%N))
                break

        moves = "".join(moves)

        if not best_move or len(moves) < len(best_move):
            best_move = moves

    return best_move

Pemakaian

solve([3, 4, 2, 1, 5])

Penjelasan

Program hanya melakukan swap semacam gelembung, menggunakan pengamatan bahwa:
S: n 1 2 3 ... n-1 0 PS: 0 2 3 ... n-1 n 1 PPS: 1 3 ... n-1 n 0 2 PPPS: 2 ... n-1 n 0 1 3 ...

Dengan kata lain, imendorong diikuti oleh swap berarti bahwa elemen iakan ditukar dengan elemen i-1, indeks membungkus. Namun, kita perlu melacak fakta bahwa daftar akan berubah dalam melakukannya.

Untuk melakukan yang lebih baik daripada sekadar semacam gelembung biasa, kami mencoba menerapkan heuristik untuk memastikan kami tidak melakukan terlalu banyak swap tidak berguna, memastikan bahwa setiap swap memindahkan kedua elemen lebih dekat ke tujuan masing-masing (saya belum secara resmi membuktikan itu benar, tetapi tampaknya berhasil sejauh ini). Kami kemudian mengambil solusi terbaik dari banyak konfigurasi (CPython membutuhkan waktu, tetapi PyPy menyelesaikan pekerjaan ini dengan relatif cepat).

Mencetak gol

Output pada Pastebin.


Bubble sort naif sebelumnya. (skor: 1126232)

Yang ini pasti berhasil.

def cyclic_sorted(nums):
    for i in range(len(nums)):
        if nums[i:] + nums[:i] == sorted(nums):
            return i
    return False

def solve(nums):
    N = len(nums)

    def gt(a, b):
        if a == 1 and b == N:
            return True
        if a == N and b == 1:
            return False
        return a > b

    moves = []
    head_ptr = 0 # Which element should be at the start of the list if we actually moved stuff

    while nums != sorted(nums):
        pops = 0
        ptr_offset = 0

        while pops < N:
            ptr = (head_ptr + pops) % N

            if (not moves or pops > 0) and gt(nums[ptr-1], nums[ptr]):
                moves.append("P"*pops + "S")
                head_ptr = (head_ptr + pops) % N

                nums[ptr-1],nums[ptr] = nums[ptr],nums[ptr-1]
                break

            pops += 1

        else:
            moves.append("P"*cyclic_sorted(nums[head_ptr:]+nums[:head_ptr]))
            break

    return "".join(moves)
Sp3000
sumber
3
Sangat rapi bahwa ini membentuk pola walaupun datanya acak. Saya mengubah P menjadi spasi untuk daftar 256 Anda (gulir di dekat bagian bawah).
Hobi Calvin
Sangat menarik bahwa nilai implementasi Anda lebih baik daripada saya meskipun, membaca dari uraian Anda, pada dasarnya kami melakukan hal yang sama. Mungkin ada beberapa perbedaan yang saya lewatkan (belum membaca kode Anda dengan saksama). ^^
ReyCharles
8

Javascript ES6 - 607.771

function popsort( iData ) {
    var sSP, iLen,
        sSP_Opt = popsortimpl( iData.slice( 0 ), 0 ),
        iOptLen = sSP_Opt.length;

    for( var n = iData.length, i = 2; i <= n; i++ )
        if( (iLen = (sSP = popsortimpl( iData.slice( 0 ), i - 1 )).length) < iOptLen ) {
            iOptLen = iLen;
            sSP_Opt = sSP;
        }

    return sSP_Opt;
}

function popsortimpl( iData, iPivIdx ) {
    function issorted()
        { return iData.every( (_,i_) => _ == i_+1 ); }

    function swap() {
        var i_ = iData[0];

        iData[0] = iData[n-1];
        iData[n-1] = i_;

        sSP.push( 'S' );
    }

    function pop() {
        iData.push( iData.shift() );
        if( --iPivIdx < 0 )
            iPivIdx += n;

        sSP.push( 'P' );
    }

    function dist2home( iIdx_, i_ ) {
        var iHome_ = (iPivIdx + i_) %n,
            iDist1_ = iIdx_ - iHome_,
            iDist2_ = iHome_ - iIdx_;

        if( iDist1_ < 0 )
            iDist1_ += n;
        if( iDist2_ < 0 )
            iDist2_ += n;

        return Math.min( iDist1_, iDist2_*Math.max( 1, n/50 ) );
    }

    var n = iData.length,
        sSP = [];

    while( !issorted() ) {
        var iDistP = Math.max( dist2home( 0, iData[0] ), dist2home( n - 1, iData[n - 1] ) ),
            iDistS = Math.max( dist2home( n - 1, iData[0] ), dist2home( 0, iData[n - 1] ) );

        if( iDistS < iDistP )
            swap();
        else pop();
    }
    return sSP.join('');
}

Menerapkan semacam gelembung "ujung terdekat" yang menggelembungkan elemen naik atau turun tergantung pada rute terpendek (gelembung atas atau bawah gelembung) ke posisi referensi.

Saya tidak percaya rutinitasnya optimal, tetapi harus cukup dekat. Sebagai bonus, ini menjalankan set lengkap vektor uji dalam milidetik.

Diperbarui untuk beralih pada titik pivot untuk meningkatkan daya saing. Saya juga memperkenalkan heuristik untuk mengimbangi fakta bahwa unsur-unsur memiliki mobilitas yang lebih rendah untuk bergerak "naik" dari daftar daripada bergerak "turun" (sebuah fenomena yang analog dengan perbedaan dalam mobilitas elektron dan lubang dalam semikonduktor). Ada perbedaan besar antara implementasi ini dan Sp3000, maka fakta bahwa kedua skor tepat 607.771 menyeramkan. Halloween seram.

Rutin sekarang membutuhkan sekitar 20 detik untuk berjalan.

Output Sampel

popsort( [2, 7, 5, 3, 4, 6, 1] )

hasil panen

PSPPSPSPPPS
COTO
sumber
4

OCaml, skor: 1136707 1136599

Sunting: Saya kacau! Saya telah membaca Sakan bertukar elemen yang berdekatan dan melewatkan elemen pertama dan terakhir bisa ditukar. Implementasi ini dibangun di atas seperti yang terlihat jelas dari evalimplementasinya. S²: Saya memperbaikinya dengan menggunakan pengamatan bahwa Anda dapat menulis apa yang saya pikir adalah SP sebagai P S. Saya juga beralih ke versi skor yang lebih baik dengan beberapa optimasi untuk membuatnya lebih lambat.

Runtime adalah 1,33 18 1,01 detik pada sistem saya menggunakan interpreter binary yang dikompilasi.

type sp = S | P

let string_of_sp = function
  | S -> "S"
  | P -> "P"

let rec eval (e : sp list) (xs : int list) : int list =
  match e, xs with
  | [], xs -> xs
  | P :: S :: e, x1 :: x2 :: xs -> eval e (x1 :: xs @ [x2])
  | P :: e, x :: xs -> eval e (xs @ [x])
  | _ -> assert false

let generate_sp_sequence xs : sp list =
  let sorted = List.sort compare xs in
  let rec generate_sp_sequence ((xs, ys) : int list * int list) acc : sp list =
    if sorted = ys @ List.rev xs
    then acc
    else 
      match ys with
      | v1 :: v2 :: ys' ->
         if v1 > v2
         then generate_sp_sequence (v2 :: xs, v1 :: ys') (S :: P :: acc)
         else generate_sp_sequence (v1 :: xs, v2 :: ys') (P :: acc)
      | [x] ->
         let ys = List.rev (x :: xs) in
         generate_sp_sequence ([], ys) (P :: acc)
      | [] ->
         assert false
  in List.rev @@ generate_sp_sequence ([], xs) []


(** Uncomment to run tests **)
(*
let test =
  QCheck.(mk_test
            ~n:1000
            ~name:"generate_sp_sequence sorting test"
            ~pp:PP.(list int)
            Arbitrary.(list ~start:2 ~stop:256 small_int)
            (fun xs ->
             Prop.assume (List.length xs >= 2);
             eval (generate_sp_sequence ([], xs))
                  xs
             = List.sort compare xs))

let _ = QCheck.run test
 *)

let score_test = ... (* not included for brevity *)

let score =
  List.map (fun xs -> generate_sp_sequence ([], xs)) score_test
  |> List.map List.length
  |> List.fold_left (+) 0

let _ =
  print_string "The score is: ";
  print_int score;
  print_newline ()

generate_sp_sequence hanya menjalankan BubbleCort dan menampilkan langkah-langkah dalam hal S dan P. Ini dapat menghasilkan langkah-langkah yang tidak perlu.

ReyCharles
sumber
4

C, skor: 607771

Ternyata solusi saya sangat mirip dengan solusi Python Sp3000 (yang menjelaskan skor identik). Saya tidak pernah menggunakan Python jadi saya tidak bisa mengatakan dengan pasti, tapi saya pikir konsepnya persis sama:

Untuk setiap angka r dalam 0..n-1, tetapkan tujuan dari array yang disortir yang diputar r kali dari posisi awal (yaitu yang berisi P muncul dengan r = P mod n), dan temukan serangkaian gerakan yang menghasilkan konfigurasi ini. Rekam urutan gerakan terpendek untuk r.
Lakukan swap hanya ketika swap mendekatkan elemen mana saja yang lebih jauh dari posisi tujuannya (melihat ke arah mana pun), atau ketika swap menggerakkan kedua elemen lebih dekat jika keduanya sama-sama jauh.

Kode

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

/* Returns true if the array is sorted starting at position 'offset' */
int is_sorted(int *array, int n, int offset)
{
    int i;
    for (i = 0; i < n && array[(offset+i)%n] == i+1; i++);
    return i == n;
}

/* Find moves that would sort the array, assuming the first element will
   be at position (offset + n/2) % n */
int get_moves(int *array, int n, int offset, int stop_after, char *moves)
{
    int tmp, posA = n-1, posB = 0, counter = 0;
    /* Repeat until sorted or reaching the maximum number of moves (previous
       best, or the worst case upper bound if no previous solution) */
    for (; counter < stop_after && !is_sorted(array, n, posB); posA = (posA+1)%n, posB = (posB+1)%n, moves[counter++] = 'P')
    {
        /* Perform a swap if it would move whichever element is further away
           closer to its target */
        if ((array[posA] - posA + offset + n) % n >
            (array[posB] - posB + offset + n) % n)
        {
            tmp = array[posA];
            array[posA] = array[posB];
            array[posB] = tmp;
            moves[counter++] = 'S';
            if (is_sorted(array, n, posB))
            {
                break;
            }
        }
    }
    return counter;
}

int *parse_input(int _argc, char **_argv)
{
    int i;
    int *array = NULL;
    if (_argc > 1)
    {
        array = malloc((_argc-1) * sizeof(int));
        _argv[1]++; /* remove initial square bracket */
        for (i = 1; i < _argc; i++)
        {
            array[i-1] = atoi(_argv[i]);
        }
    }
    return array;
}

int main(int _argc, char **_argv)
{
    int *orig_array = parse_input(_argc, _argv);
    if (orig_array != NULL)
    {
        int n = _argc - 1;
        /* Safe worst case upper bound representing n^2 "SP" moves */
        int worst_move_count = 2*n*n;

        int i, move_count, offset;
        int best_offset, best_move_count = worst_move_count;

        char *best_moves = malloc(worst_move_count + 1);
        char *moves = malloc(worst_move_count + 1);
        int *array = malloc(n * sizeof(int));

        memset(best_moves, 0, worst_move_count + 1);
        /* Try all offsets from 0 to n-1; i.e. find solutions for:
           1, 2, 3, ..., n-2, n-1, n
           2, 3, 4, ..., n-1, n,   1
           3, 4, 5, ..., n,   1,   2
           ...
           n, 1, 2, ..., n-3, n-2, n-1
        */
        for (offset = 0; offset < n; offset++)
        {
            memset(moves, 0, worst_move_count + 1);
            memcpy(array, orig_array, n*sizeof(int));
            move_count = get_moves(array, n, offset, best_move_count, moves);
            if (move_count < best_move_count)
            {
                strcpy(best_moves, moves);
                best_move_count = move_count;
                best_offset = offset;
            }
        }
        printf("%s\n", best_moves);

        free(best_moves);
        free(moves);
        free(array);
        free(orig_array);
    }
    return 0;
}

Pemakaian

$ gcc -Ofast -o popswap popswap.c
$ ./popswap [2, 7, 5, 3, 4, 6, 1]
PSPPSPSPPPS
$ ./popswap [2, 7, 5, 3, 4, 6, 1]|tr -d '\n'|wc -c
11

Mencetak gol

Naskah:

#!/bin/sh
grep ^\\[ input.txt|while read line; do
    len=`echo $line|wc -w`;
    score=`./popswap $line|tr -d '\n'|wc -c`;
    echo "    $len $score";
done

Hasil:

$ ./scoring.sh |cut -d' ' -f 6|paste -sd+|bc
607771
$ time ./scoring.sh > /dev/null

real    0m1.926s
user    0m1.879s
sys     0m0.122s
$ ./scoring.sh
3 2
7 11
41 893
52 1546
65 2543
69 2686
75 3391
80 3728
103 6483
108 0
109 7103
151 14371
173 18663
177 19728
192 22824
201 24828
201 24752
205 25573
211 27509
226 32221
227 31928
230 33630
238 34603
241 36676
241 36505
244 37759
246 38106
250 38264
253 40040
256 41405
Mike Patterson
sumber
1

C ++ 11, skor: 1122052

Implementasi yang sangat naif yang tidak memperhitungkan lubang dalam urutan input tetapi memperhitungkan nilai minimal dalam urutan.

Daftar argumen hanyalah urutan angka, tanpa koma dan tanda kurung (mis: "./a.out 2 7 5")

#include <iostream>
#include <cstring>
#include <string>
using namespace std;

int* input  = 0;
int s_count = 0;
int p_count = 0;
int len;
int last;

void do_swap()
{
  int a = input[0];
  input[0] = input[last];
  input[last] = a;
  s_count++;
  cerr << "S";
}

void do_pop()
{
  int a = input[0];
  memmove(&input[0], &input[1], sizeof(int) * last);
  input[last] = a;
  p_count++;
  cerr << "P";
}

bool is_sorted()
{
  int i = 0;
  while (input[i] < input[i+1]) { if (++i == last) return true; }
  return false;
}

int main(int argc, char** args)
{
  int min_val = 256;

  len = argc - 1;
  last = len - 1;
  if (len <= 1)
    return 1;

  input = new int[len];
  for (int i = 2; i <= argc; i++)
  {
    int j = stoi(args[i-1]);
    input[i-2] = j;
    min_val = min(min_val, j);
  }

  while (!is_sorted())
  {
    if ((input[0]) == min_val)
      do_pop();
    else if (input[0] < input[last])
      do_swap();
    else if (input[0] > input[last])
      do_pop();
  }

  cerr << endl;
  cout << len << " " << (s_count + p_count) << endl;
  delete [] input;
  return 0;
}

Mencetak gol

3 2
7 19
41 1642
52 3236
65 4787
69 4966
75 6644
80 6090
103 12142
108 0
109 13774
151 24976
173 34061
177 37835
192 41994
201 48316
201 47276
205 49439
211 51551
226 58859
227 56824
230 58994
238 64227
241 65774
241 66287
244 71345
246 73426
250 71766
253 68629
256 77171
BlakBat
sumber